Let $E$ be an elliptic curve over field $\mathbb{Z}/p\mathbb{Z}$.
The curve group $E(\mathbb{Z}/p\mathbb{Z})$ is always a) cyclic or b) direct product of two cyclic groups.
First question: How do I tell for a given field and curve if it's the case a) or b) ?
Another question follows from an example: The curve $E: y^2 = x^3 - x$ over $\mathbb{Z}/71\mathbb{Z}$ has 71 affine points and a point at infinity so it's group is isomorphic to $\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/36\mathbb{Z}$ (that's the case b)).
Second question is: Why it's not isomorphic to $\mathbb{Z}/3\mathbb{Z}\times\mathbb{Z}/24\mathbb{Z}$ or $\mathbb{Z}/4\mathbb{Z}\times\mathbb{Z}/18\mathbb{Z}$ ?
In other words, for b) case curves, how do I tell which are those two in the product?
For simplicity, I’ll write $C_n$ for $\mathbb Z/n\mathbb Z$.
First, I think you’ll find that $C_2\times C_{36}$ and $C_4\times C_{18}$ are isomorphic. The best way to write something like this is to make sure the indices divide: $2|36$, but $4$ does not divide $18$.
Your question about why $C_3\times C_{24}$ does not occur is much more interesting, and I’m almost out in the water over my head in attempting an answer. There’s a fancy gadget called the $e_n$-pairing on the points of order $n$ of an elliptic curve, and it’s defined over the ground field, with values in the multiplicative group scheme $\mathbf G_{\mathrm m}$. The upshot is that if all $n^2$ points of the elliptic curve of period $n$ are $k$-rational ($k$ being the ground field), then the $n$-th roots of unity have to be in $k$ as well. But of course $\mathbb F_{71}$ doesn’t have cube roots of unity. (I know that I’ve left out far too many details here, but the topic is fairly advanced, and I don’t have the skills.)